Download Grade 6 Math Circles Pythagorean Theorem

Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Centre for Education in
Mathematics and Computing
Grade 6 Math Circles
March 24/25, 2015
Pythagorean Theorem
Triangles: They’re Alright When They’re All Right
Today, we’re going to talk about a specific theorem that has to do with right-angled triangles. Triangles are arguably one of the most important fundamental shapes, besides circles.
We’re going to take a look at a proof of the theorem and we’re going to apply the theorem
in problems.
Warm-Up
Try to answer the following 4 questions in 2 minutes.
1. What is the length of the hypotenuse if the legs are 8 and 15?
2. What is the length of the missing leg if the hypotenuse is 25 and one leg is 7?
3. If the length of a rectangle is 16 cm, and diagonal is 34 cm, what is the width?
4. Using Euclid’s Formula where n is 36 and m is 75, find a Pythagorean Triple.
Terminology
The Pythagorean Theorem can only be used in right-angled triangles. A right-angled triangle
is a triangle with a 90◦ angle (a right angle).
The hypotenuse is the longest side of the right-angled triangle. It is opposite the right angle.
The legs of a triangle are the two other sides of a right-angled triangle.
A vertex is any corner of the triangle.
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Try it out:
Label the diagram below.
The Theorem
The Pythagorean Theorem states that a2 + b2 = c2 , where a and b are the legs and c is the
hypotenuse.
c
b
a
Try it out:
Solve for the missing lengths.
1. a = 1, b =
√
3, c =?
2. a = 6, b =?, c = 10
3. a =?, b = 24, c = 25
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A Visualization of the Theorem
5
4
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Examine the right-angled triangle above, with hypotenuse 5, and legs of 3 and 4. You can
place squares on each of the sides. The hypotenuse has a 5 × 5 square, and the legs have
3 × 3 and 4 × 4 squares respectively.
This means that the hypotenuse has 52 = 25 unit (1 × 1) squares, and the legs have 32 = 9
and 42 = 16 unit squares respectively.
Let’s see if our two smaller squares fit into the bigger one:
Now, it doesn’t look like we could fit our 3 × 3 square in there; but notice that there are
9 orange 1 × 1 squares left, which is the same amount of red 1 × 1 squares that make up
our 3 × 3 square! Let’s break apart our red 3 × 3 square and fit it into the rest of our
5 × 5 square above.
Well, it seems like the squares of the two smaller sides of the right triangle can be used to
make up the square of the biggest side. In particular, 32 + 42 = 52 .
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A Proof of the Theorem
We will prove this theorem by using pictures. Let’s arrange 4 of the same right triangles to
get the figure below:
Well, if the 4 side lengths of the figure in the middle are all c, this tells us that we have
a square. The shaded region below represents the area of the square in the middle of our
figure. This area is equal to c2 .
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Now, remember that we need to show that a2 + b2 = c2 . So, because we can, let’s play
around with the arrangement of the 4 right triangles in the previous figure:
Notice that the total area of our figure did not change, because all we did was move the right
triangles around. Then, since the area of the 4 right triangles has also not changed, this
tells us that the area of our two shaded regions must equal the area of our original shaded
region, c2 . But, the area of the smaller shaded region is just a2 , and the area of the larger
shaded region is just b2 .
Therefore, we conclude that a2 + b2 = c2 .
Pythagorean Triples
A Pythagorean Triple is a set of 3 numbers that form a perfect solution to the Pythagorean
Theorem. The easiest triple to think of is
345
We can take any multiple of this triple (multiply all 3 numbers by the same number) and
still have the Pythagorean Theorem work properly. Multiples of this triple are:
6 8 10
9 12 15
12 16 20
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There are also other Pythagorean Triples that are not multiples of the triple 3 4 5.
5 12 13
7 24 25
9 40 41
11 60 61
Do you notice anything about these triples?
1. The 2nd two numbers only differ by 1.
2. The sum of the 2nd two numbers is equal to the square of the first number.
a
b
c
a2
a2
2
3
4
5
9
4.5
5
12 13
25
12.5
7
24 25
49
24.5
9
40 41
81
40.5
11 60 61 121 60.5
Try it out: Form a Pythagorean Triple with 13 as the first number.
Another way to find Pythagorean Triples directly is by using Euclid’s Formula.
The formula requires any two positive integers, m and n, where m > n.
a = m2 − n2
b = 2mn
c = m 2 + n2
Proof:
a2 + b2 = (m2 − n2 )2 + (2mn)2 = (m2 + n2 )2 = c2
Try it out: Form a Pythagorean Triple where m is 18 and n is 11.
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Wrap-Up
Today, you looked at a very important theorem involving right-angled triangles. You will
discover that the Pythagorean Theorem is extremely useful in a variety of physical situations
and most geometry problems involving triangles.
Problem Set
Complete the following 15 problems. You may use a calculator.
1. Solve for the missing length, assuming c is the hypotenuse. Leave answers in radical
√
form where necessary ( number).
(a) a = 2, b = 2, c =?
(b) a = 2, b = 4, c =?
(c) a = 84, b =?, c = 300
(d) a =?, b = 404, c = 505
(e) a = 28, b =?, c = 100
2. Find the Pythagorean Triple that contains a first number of 17.
3. Find the Pythagorean Triple where m is 72 and n is 14.
4. Which of the following is NOT a Pythagorean Triple?
(a) 19 180 181
(b) 35 612 613
(c) 40 76 86
(d) 12 35 37
5. Which of the following is a Pythagorean Triple?
(a) 21 24 33
(b) 10 24 25
(c) 3 3 6
(d) 22 120 122
6. If a bird is on the ground, 20 metres away from the base of a building, and the bird
flies 29 metres to the top of the building, how tall is the building?
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7. Harry is on his broom 18 metres above the point he took off from. Ron is standing
on the ground, 24 metres away from Harry’s takeoff point. How far apart are Harry
and Ron?
8. Solve for the missing lengths.
(a) Find x.
(b) Find x.
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(c) Find x.
9. Solve for the length of AB
9
10. Solve for the area of the following shapes:
(a) test
(b) test
(c) test
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11. Given the rectangular prism below, solve for the length of AH.
12. Triangle ABC below is right-angled with AB = 10 and AC = 8. If BC = 3DC, what
is the length of AD?
B
10
D
C
A
8
13. If the area of an isosceles triangle is 8 cm2 , and the base and height of the triangle are
the same, what are the lengths of the other two sides?
14. Calculate the value of h in the diagram below.
20
h
25
Don’t forget question 15 on the next page!
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15. Solve for the area of the trapezoid given the following information (round any calculations to one decimal place).
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